3.168 \(\int \frac{(c e+d e x)^4}{(a+b \sinh ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=320 \[ \frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}-\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac{27 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac{25 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^4 \sqrt{(c+d x)^2+1} (c+d x)^4}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]

[Out]

-(e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(2*b*d*(a + b*ArcSinh[c + d*x])^2) - (2*e^4*(c + d*x)^3)/(b^2*d*(a +
b*ArcSinh[c + d*x])) - (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcSinh[c + d*x])) + (e^4*Cosh[a/b]*CoshIntegral[(a
+ b*ArcSinh[c + d*x])/b])/(16*b^3*d) - (27*e^4*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c + d*x]))/b])/(3
2*b^3*d) + (25*e^4*Cosh[(5*a)/b]*CoshIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(32*b^3*d) - (e^4*Sinh[a/b]*Sin
hIntegral[(a + b*ArcSinh[c + d*x])/b])/(16*b^3*d) + (27*e^4*Sinh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c + d
*x]))/b])/(32*b^3*d) - (25*e^4*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c + d*x]))/b])/(32*b^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.891697, antiderivative size = 316, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3303, 3298, 3301} \[ \frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{16 b^3 d}-\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{16 b^3 d}+\frac{27 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{25 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^4 \sqrt{(c+d x)^2+1} (c+d x)^4}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

-(e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(2*b*d*(a + b*ArcSinh[c + d*x])^2) - (2*e^4*(c + d*x)^3)/(b^2*d*(a +
b*ArcSinh[c + d*x])) - (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcSinh[c + d*x])) + (e^4*Cosh[a/b]*CoshIntegral[a/
b + ArcSinh[c + d*x]])/(16*b^3*d) - (27*e^4*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(32*b^3*
d) + (25*e^4*Cosh[(5*a)/b]*CoshIntegral[(5*a)/b + 5*ArcSinh[c + d*x]])/(32*b^3*d) - (e^4*Sinh[a/b]*SinhIntegra
l[a/b + ArcSinh[c + d*x]])/(16*b^3*d) + (27*e^4*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c + d*x]])/(32*
b^3*d) - (25*e^4*Sinh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c + d*x]])/(32*b^3*d)

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5667

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c
^2*x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{\left (2 e^4\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}+\frac{\left (5 e^4\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{x^4}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^4(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (6 e^4\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \left (\frac{\cosh (x)}{8 (a+b x)}-\frac{3 \cosh (3 x)}{16 (a+b x)}+\frac{\cosh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}-\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (25 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{\left (75 e^4\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{\left (3 e^4 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (25 e^4 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{\left (3 e^4 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (75 e^4 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}+\frac{\left (25 e^4 \cosh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}+\frac{\left (3 e^4 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (25 e^4 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{\left (3 e^4 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (75 e^4 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}-\frac{\left (25 e^4 \sinh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e^4 (c+d x)^3}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^4 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{16 b^3 d}-\frac{27 e^4 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}+\frac{25 e^4 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{e^4 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{16 b^3 d}+\frac{27 e^4 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}-\frac{25 e^4 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c+d x)\right )}{32 b^3 d}\\ \end{align*}

Mathematica [A]  time = 1.23823, size = 316, normalized size = 0.99 \[ \frac{e^4 \left (-\frac{16 b^2 \sqrt{(c+d x)^2+1} (c+d x)^4}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+48 \left (-\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+25 \left (2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+\frac{16 b \left (-5 (c+d x)^5-4 (c+d x)^3\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{32 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(e^4*((-16*b^2*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^2 + (16*b*(-4*(c + d*x)^3 - 5*(c +
d*x)^5))/(a + b*ArcSinh[c + d*x]) + 48*(-(Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]]) + Cosh[(3*a)/b]*Cosh
Integral[3*(a/b + ArcSinh[c + d*x])] + Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]] - Sinh[(3*a)/b]*SinhInte
gral[3*(a/b + ArcSinh[c + d*x])]) + 25*(2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c + d*x]] - 3*Cosh[(3*a)/b]*Cos
hIntegral[3*(a/b + ArcSinh[c + d*x])] + Cosh[(5*a)/b]*CoshIntegral[5*(a/b + ArcSinh[c + d*x])] - 2*Sinh[a/b]*S
inhIntegral[a/b + ArcSinh[c + d*x]] + 3*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c + d*x])] - Sinh[(5*a)/b]
*SinhIntegral[5*(a/b + ArcSinh[c + d*x])])))/(32*b^3*d)

________________________________________________________________________________________

Maple [B]  time = 0.224, size = 896, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x)

[Out]

1/d*(-1/64*(16*(d*x+c)^5-16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+20*(d*x+c)^3-12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+5*d*x+
5*c-(1+(d*x+c)^2)^(1/2))*e^4*(5*b*arcsinh(d*x+c)+5*a-b)/b^2/(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)+a^2)-25
/64*e^4/b^3*exp(5*a/b)*Ei(1,5*arcsinh(d*x+c)+5*a/b)+3/64*(4*(d*x+c)^3-4*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+3*d*x+3*
c-(1+(d*x+c)^2)^(1/2))*e^4*(3*b*arcsinh(d*x+c)+3*a-b)/b^2/(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)+a^2)+27/6
4*e^4/b^3*exp(3*a/b)*Ei(1,3*arcsinh(d*x+c)+3*a/b)-1/32*(-(1+(d*x+c)^2)^(1/2)+d*x+c)*e^4*(b*arcsinh(d*x+c)+a-b)
/b^2/(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)+a^2)-1/32*e^4/b^3*exp(a/b)*Ei(1,arcsinh(d*x+c)+a/b)-1/32*e^4/b
*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-1/32*e^4/b^2*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+
c))-1/32*e^4/b^3*exp(-a/b)*Ei(1,-arcsinh(d*x+c)-a/b)+3/64*e^4/b*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c)^
2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2+9/64*e^4/b^2*(4*(d*x+c)^3+3*d*x+3*c+4*(d*x+c)^2*(1+(d*x+c
)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))+27/64*e^4/b^3*exp(-3*a/b)*Ei(1,-3*arcsinh(d*x+c)-3*a/b)-1
/64*e^4/b*(16*(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/
2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-5/64*e^4/b^2*(16*(d*x+c)^5+20*(d*x+c)^3+16*(d*x+c)^4*(1+(d*x+c)
^2)^(1/2)+5*d*x+5*c+12*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-25/64*e^4/b^3*e
xp(-5*a/b)*Ei(1,-5*arcsinh(d*x+c)-5*a/b))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{arsinh}\left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4)/(b^3*arcsinh(d*x + c)^3
+ 3*a*b^2*arcsinh(d*x + c)^2 + 3*a^2*b*arcsinh(d*x + c) + a^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asinh(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{4}}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^4/(b*arcsinh(d*x + c) + a)^3, x)